4a Three point charges lie along the x axis, The positive charge q115.0C is at x 2.00 m, the positive charge q2 6.00 C is at the origin, and the net force acting on q3 is zero. What is the x coordinate of q3?

b .Find the net force on q2 and its direction

1 answer

a) To find the position of q3, we know that the net force on it is zero. This means that the forces exerted on q3 by q1 and q2 must cancel each other out.

We can use Coulomb's Law to calculate the force between q1 and q3:

F1 = k * (q1 * q3) / r^2

where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2), q1 and q3 are the charges, and r is the distance between them.

Similarly, we can calculate the force between q2 and q3:

F2 = k * (q2 * q3) / r^2

Since the net force on q3 is zero, F1 + F2 = 0.

Thus, k * (q1 * q3) / r^2 + k * (q2 * q3) / r^2 = 0

Simplifying, we get:

q1 * q3 + q2 * q3 = 0

(115C)(q3) + (6C)(q3) = 0

Solving this equation, we find:

121C * q3 = 0

Since the product of the charges is zero, one of the charges (q3) must be zero. This means that there is no charge q3, so it does not have an x-coordinate.

b) To find the net force on q2, we need to calculate the forces exerted on it by q1 and q3.

The force between q1 and q2 can be calculated using Coulomb's Law:

F1 = k * (q1 * q2) / r^2

where q1 and q2 are the charges and r is the distance between them.

In this case, q1 = 115C, q2 = 6C, and r = 2.00m.

Using the value of k = 9 x 10^9 Nm^2/C^2, we can calculate F1:

F1 = (9 x 10^9 Nm^2/C^2) * (115C * 6C) / (2.00m)^2

Simplifying, we find:

F1 = 3.4375 x 10^11 N

The force between q3 and q2 is zero since q3 does not exist. Therefore, the net force on q2 is equal to the force exerted on it by q1, which is 3.4375 x 10^11 N. The direction of this force will be towards q1 (in the negative x-direction), since q1 is a positive charge and q2 is a negative charge.